武汉大学第十六届程序设计大赛(网络赛) F.A-maze-ing(Tarjan，缩点，求最长链)

                             A-maze-ing

题目传送门：滴滴滴

Long long age, a wise but barbaric king made a plan to convert the
arena to a maze so that he could get pleasure from watching his
servant running the maze confusedly. The structure of the arena could
be abstracted into a directed connected graph comprised of n (1 ≤ n ≤
105 ) nodes and m (1 ≤ m ≤ 2 × 105 ) edges. The king had not decided
where to set up the starting nodes and the end node. So the king
proposed a requirement. Whichever two points u, v he chose, there
existed a path from one point to the other(a path from u to v or from
v to u). Moreover, the king hoped to improve the difficulty of the
game, so the number of nodes in the maze should be as large as
possible. You are only required to output the maximum number of nodes
in the maze.

Input

There are two positive integers n and m in the first line. And then m
lines follow, in each line there are two positive integers u and v,
describing that an edge from node u to node v.

Output

Just output one integer, the maximum size of the maze

Sample Input

3 2
1 2
1 3

Sample Output

2



题目意思： n个点m条边,求出最长的一条连接长度，比如样例3-1-2长为2
比赛的时候慌的一匹，默默翻翻字典。后面比赛完，问了下学长emmm。
思路：tarjan缩点，在重新建图，记录下每个连通图的点数，然后topsort求最长链就行了

#include<iostream>  
#include<cstdio>  
#include<cstring>  
#include<ctime>  
#include<cstdlib>  
#include<algorithm>  
#include<cmath>  
#include<string>  
#include<queue>  
#include<vector>  
#include<stack>  
#include<list>  
#include<map>  
#include<set>  
#define P pair<int ,int >  
using namespace std;  
const int maxn = 100000+5;  
int Laxt[maxn],cnt,n,m,ans;  
int times,dfn[maxn],low[maxn],scc[maxn],scc_cnt,sccnum[maxn],in[maxn],dis[maxn];  
int stc[maxn],top,instc[maxn];  
map<P,int> mp;  
struct node  
{  
    int from;  
    int to;  
    int next;  
    int w;  
} edge[maxn*20],e[maxn*20] ;  
void add(int u,int v,int w)  
{  
    edge[++cnt].from=u;  
    edge[cnt].next=Laxt[u];  
    Laxt[u]=cnt;  
    edge[cnt].to=v;  
    edge[cnt].w=w;  
}  
int dfs(int u)  
{  
    dfn[u]=low[u]=++times;  
    stc[++top]=u;  
    instc[u]=1;  
    for(int i=Laxt[u]; i; i=edge[i].next)  
    {  
        int v=edge[i].to;  
        if(!dfn[v])  
        {  
            dfs(v);  
            low[u]=min(low[u],low[v]);  
        }  
        else if(instc[v])  
        {  
            low[u]=min(low[u],low[v]);  
        }  
    }  
    if(dfn[u]==low[u])  
    {  
        scc_cnt++;  
        while(true)  
        {  
            int x=stc[top--];  
            scc[x]=scc_cnt;  
            sccnum[scc_cnt]++;  
            instc[x]=0;  
            if(x==u) break;  
        }  
    }  
}  
queue<int> q;  
void tarjan()  
{  
    for(int i=1; i<=n; i++)  
        if(!dfn[i]) dfs(i);  
    for(int i=1; i<=cnt; i++)  
    {  
        e[i]=edge[i];  
    }  
    memset(Laxt,0,sizeof(Laxt));  
    for(int i=1; i<=cnt; i++)  
    {  
        if(scc[e[i].from]!=scc[e[i].to]&&!mp[make_pair(scc[e[i].from],scc[e[i].to])])  
        {  
            in[scc[e[i].to]]++;  
            mp[make_pair(scc[e[i].from],scc[e[i].to])]=1;  
            add(scc[e[i].from],scc[e[i].to],1);  
        }  
    }  
    for(int i=1; i<=scc_cnt; i++)  
    {  
        if(in[i]==0)  
        {  
            q.push(i);  
            dis[i]=sccnum[i];  
        }  
        else  
        {  
            dis[i]=0;  
        }  
    }  
}  
void topsort()  
{  
    int now;  
    while(!q.empty())  
    {  
        now=q.front();  
        q.pop();  
        for(int j=Laxt[now]; j; j=edge[j].next)  
        {  
            in[edge[j].to]--;  
            if(in[edge[j].to]==0)q.push(edge[j].to);  
            if(dis[now]+sccnum[edge[j].to]>dis[edge[j].to])  
            {  
                dis[edge[j].to]=dis[now]+sccnum[edge[j].to];  
            }  
        }  
    }  
}  
int main()  
{  
    scanf("%d %d",&n,&m);  
        int a,b;  
        for(int i=1; i<=m; i++)  
        {  
            scanf("%d %d",&a,&b);  
            add(a,b,1);  
        }  
        tarjan();  
        topsort();  
        int maxnode=0;  
        for(int i=1; i<=n; i++)  
        {  
            if(maxnode<dis[i])  
            {  
                maxnode=dis[i];  
            }  
        }  
        printf("%d
",maxnode);  
}  

打比赛时候有思路是不可能的，这辈子不可能的，只可能补题来维持生活这样子。

PS：摸鱼怪的博客分享，欢迎感谢各路大牛的指点~